This is the error :
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in test/u_p_s_author.php on line 49
Did I miss anything?
QUOTE(dontKnowJava @ 29 Jan, 2008 - 11:49 AM)
try this
CODE
$connect = mysql_connect("database","username","password") or die(mysql_error());
$select = mysql_select_db("database") or die(mysql_error());
That's not a problem: The problem is the select command
QUOTE(baavgai @ 29 Jan, 2008 - 12:09 PM)
A couple things. I'm not sure you can use "tableName as a"; I believe you have to drop the "as" for that.
The
"... a.Author_Name = 'upper($field) LIKE'%$find%''" looks odd and probably wont do what you're looking for. Certainly won't work in SQL.
Here's some syntatically correct SQL. See if you can take it from there.
CODE
SELECT p.Author_ID, a.Author_Name
from PUBLICATION p
inner join AUTHORS a
on p.Author_ID=a.Author_ID
where upper(a.Author_Name) like '%KIN%'
Hope this helps.
Thanks. But I want to view all authors in the database without select one name after the other. If I have 10 authors, I want select 5 authors with different publication at the same time not selecting them one by one. NO.
Thanks.
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